Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Posted on December 4, 2010 by The Mathematician

Clever student:

I know!

$x^{0}$ = $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$ .

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$ = $0^{1+x-1}$ = $0^{1} \times 0^{x-1}$ = $0 \times 0^{x-1}$ = $0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$ .

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

= $\lim_{x \to 0^{+}} \exp(x \log(x))$

= $\exp( \lim_{x \to 0^{+} } x \log(x) )$

= $\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

= $\exp( \lim_{x \to 0^{+} } -x )$

= $\exp( 0)$

= $1$

So, since $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

High School Teacher:

Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$ . The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

Calculus Teacher:

For all $x>0$ , we have

$0^{x} = 0$ .

Hence,

$\lim_{x \to 0^{+}} 0^{x} = 0$

That is, as x gets arbitrarily close to $0$ (but remains positive), $0^{x}$ stays at $0$ .

On the other hand, for real numbers y such that $y \ne 0$ , we have that

$y^{0} = 1$ .

Hence,

$\lim_{y \to 0} y^{0} = 1$

That is, as y gets arbitrarily close to $0$ , $y^{0}$ stays at $1$ .

Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$ . In particular, when we approach (0,0) along the line with x=0 we get

$\lim_{y \to 0} f(0,y) = 1$

but when we approach (0,0) along the line segment with y=0 and x>0 we get

$\lim_{x \to 0^{+}} f(x,0) = 0$ .

Therefore, the value of $\lim_{(x,y) \to (0,0)} y^{x}$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$ .

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

$y^x$ := $1 \times y \times y \cdots \times y$

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

$y^{x}$ = $1 \times y$ .

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

$y^{0}$ = $1$

which holds for any y. Hence, when y is zero, we have

$0^0 = 1$ .

Look, we’ve just proved that $0^0 = 1$ ! But this is only for one possible definition of $y^x$ . What if we used another definition? For example, suppose that we decide to define $y^x$ as

$y^x$ := $\lim_{z \to x^{+}} y^{z}$ .

In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use $y^z$ in our definition of $y^x$ , which seems to be recursive. The reason it is okay is because we are working here only with $z>0$ , and everyone agrees about what $y^z$ equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

$0^0$ = $\lim_{x \to 0^{+}} 0^{x}$ = $\lim_{x \to 0^{+}} 0$ = $0$

Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$ . Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for x>0 and y>0 we know what we mean by $y^x$ . But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that $0^0=1$ ? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined. For example, consider the binomial theorem, which says that:

$(a+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$

where $\binom{x}{k}$ means the binomial coefficients.

Now, setting a=0 on both sides and assuming $b \ne 0$ we get

$b^x$

= $(0+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}$

= $\binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots$

= $\binom{x}{0} 0^0 b^{x}$

= $0^0 b^{x}$

where, I’ve used that $0^k = 0$ for k>0, and that $\binom{x}{0} = 1$ . Now, it so happens that the right hand side has the magical factor $0^0$ . Hence, if we do not use $0^0 = 1$ then the binomial theorem (as written) does not hold when a=0 because then $b^x$ does not equal $0^0 b^{x}$ .

If mathematicians were to use $0^0 = 0$ , or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using $0^0 = 1$ .

There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

This entry was posted in -- By the Mathematician, Math, Philosophical. Bookmark the permalink.

1,177 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Yee says:

February 15, 2013 at 6:27 pm

The Cool Dude,
Undefined is a definition.
The Cool Dude says:

February 16, 2013 at 1:07 pm

I have no reply to that, so instead I will reply with the work I have done on proving ln(0)=1/0.

First and foremost, ln(x) is outright defined as
lim(a→0) of (x^a-1)/a
The natural log function is defined by this limit for all of its points, which means the limit equation must hold true for all scenarios.
For ln(0),
x=0
(0^a-1)/a, for negative a, is consistently 1/0
(0^a-1)/a, for positive a, is -1/a (a→0), which is again, 1/0.

One might argue that this makes ln(0) negative infinity, but that really is just a matter of opinion, as limits leave room for error by a difference of an infinitesimal. ln(0) is certainly one of the two.
Yee says:

February 17, 2013 at 1:27 am

The Cool Dude,
You have digressed.
If you want to discuss ln(0),
you can start a new topic.
The Cool Dude says:

February 17, 2013 at 10:10 am

Yee,
It is only a digression if it is irrelevant, or does not add new ideas.
The natural log function is relevant to exponents.
The topic is 0^0.
I am about to add new ideas to the conversation.

Referring to a method of definition of 0^0 I proposed earlier, was denied by my assumption of ln(0)=1/0.
Now that I can properly define ln(0) as 1/0, the method of definition can be reevaluated.

If ln(0)=1/0, then e^(1/0) has a possible solution of 0.
If e^(1/0)=0, then e^(0*1/0)=0^0, or that, e^(0/0)=0^0, and that ln(0^0)=0/0.

With ln(0/0)=0*ln(0),
ln(0/0)=0/0,
And e^(0/0)=0/0,
Then with our prior conclusion, 0^0=0/0.

The ability to use our topic entity loosely is the key to defining it further.
jojo says:

February 17, 2013 at 11:29 pm

Of course 0^0 power is undefined when using limits as the values never actually reach zero they become arbitrarily close to it, and thus depending on how you approach the limit 0^0 can have an infinite number of values between 0 and 1 inclusive. However when the values of X^Y are zero, not simply arbitrarily close to zero (an important distinction), then the value of X^Y must be one.
Yee says:

February 18, 2013 at 9:20 am

The Cool Dude,
ln(0^0) does not equal 0*ln(0)
The discussion of ln(0) is digressive.
The Cool Dude says:

February 19, 2013 at 6:31 am

Yee,
I did not directly define ln(0^0) as 0*ln(0).
I defined e^(1/0) as equaling 0, and raised both to the power of zero.
Yee says:

February 19, 2013 at 5:49 pm

The Cool Dude,
1/0 is undefined.
How can you define e^(1/0) as 0?
The Cool Dude says:

February 20, 2013 at 5:01 pm

Yee,
1/0 can easily be defined in the same way that √(-1) can.
We define √(-1), i, as the solution to i^2=-1.
In the same way, we can define 1/0 as the solution, U, to 1/U=0.

As for e^(1/0), if ln(0)=1/0, then a possible solution of e^(1/0) is zero.
Though even considering the alternate definition of e^ln(0) as 1/0, the product is the same after raising both sides to the power of zero.
Yee says:

February 20, 2013 at 6:49 pm

The Cool Dude,
You have digressed.
The Wonderer says:

February 20, 2013 at 7:38 pm

You cant define 1/o as U. It wont work due to how easy it can be manipulated.
The Cool Dude says:

February 21, 2013 at 3:29 pm

Yee,
That really is a matter of opinion. I feel as though I am expanding the definition of 0^0. Given the topic, it could not be any more progressive than it already is.

The Wonderer
Dividing by zero does not causes incorrect statements, only when simplifying or expanding improperly do you encounter inconsistencies. If divide by zero, but do not multiply by zero, or raise to the power of zero, you will never come to an incorrect conclusion. The exact same problem occurs when solving certain calculus problems. Often, the apparent result would be 0/0, but since Calculus consistently refers to limits, simplifying an expression of 0/0 to 1 would likely be incorrect.

Mathematics is only really composed of two methods of progress: operation, and replacement. Operation is rarely done improperly, mostly by mathematical beginners and students. Replacement on the other hand, (Simplification, expansion, etc) is done incorrectly by many mathematicians, regardless of their experience. Certain replacements would seem logical to make, like when replacing x^0 with 1, since in the entire finite complex plane, there is only one value that does not comply with the replacement. In the same way, it would be even more logical to replace 0*x with 0, since there are no values at all in the finite complex plane that do not comply, though when x is the reciprocal of zero, then you again have an inconsistency.
The Wonderer says:

February 21, 2013 at 9:34 pm

The cool dude. Once you expressed ‘Dividing by zero does not causes incorrect statements, only when simplifying or expanding improperly do you encounter inconsistencies. If divide by zero, but do not multiply by zero, or raise to the power of zero, you will never come to an incorrect conclusion.’ I was thinking of multiplying by zero. I had recently looked at another question on this site that has the same exact question of can you define 1/0 as Q I believe he said. The link to the post is – https://www.askamathematician.com/2011/06/q-can-you-fix-the-10-problem-by-defining-10-as-a-new-number/
Yee says:

February 21, 2013 at 11:20 pm

The Cool Dude,

0 can not be a denominator.
The Cool Dude says:

February 22, 2013 at 5:09 pm

The Wonderer
In that link, it was attempted that 1/0 would be defined as the solution Q to 0*Q=1. Defining it that way is literally exactly what I said causes problems.
If you multiply and divide by zero at the same time, you have created a factor of 0/0.
Multiplying and dividing by zero is not wrong, however, it is incorrect to simplify 0/0 to 1, for all the many reasons explained in that post, which is exactly why I did not define it in that way.

Instead of defining U as the solution to 0*U=1, instead define U as the solution to 1/U=0. Any operation made to attempt to rewrite the equation to cause an incorrect statement would first create a 0/0 somewhere, which again, can only be make the equation incorrect by simplifying to 1.

Yee,
Operation is never incorrect. It’s how we transform what we have operated on in response that causes problems. I see no reason why dividing by zero is a problem. What I see is an action that is associated to a problem.

Believing division by zero is wrong due to incorrect solutions is an association fallacy.
The real problem is simplifying 0/0 to 1.

http://en.wikipedia.org/wiki/Association_fallacy
The Cool Dude says:

February 22, 2013 at 5:32 pm

Further on the idea of defining U as the solution to 0*U=1, which does in fact apply to the topic conversation, if U were to be defined as the solution 0*U=1, it would also have to be the solution to 0*U=n, for all n*0=0, since multiplying the first equation by n does not change the left side, as the n*0 is defined as equaling zero, it simply becomes
0*U=n
For all n*0=0.

Thus, if 1/0 were defined in this way, it would further support the idea that 0/0 is a complete set of all real and imaginary numbers. Defining 0*U=1 is technically correct, but assuming that 0*U is only 1 is not.
Yee says:

February 23, 2013 at 9:19 am

The Cool Dude,
The problem of dividing by 0 has been discussed in many topics.
It is not necessary to discuss it again here.
The Cool Dude says:

February 24, 2013 at 2:41 pm

Yee,
Division by zero is relevant to 0/0, which is relevant to the topic of conversation, 0^0.

Reevaluating old topics is important to progression in any subject.
Everyone on Earth used to believe that flies spawned from meat.
Everyone on Earth was wrong.
Yee says:

February 24, 2013 at 6:05 pm

The Cool Dude,
0^0 is irrelevant to 0/0.
The Cool Dude says:

February 25, 2013 at 6:33 am

Yee,
Considering that if index laws were used, they would literally be the exact same thing, I’m led to believe that the two are extremely relevant to each other.

Further, I’ve already shown that ln(0)=1/0, by definition of the natural log function, and why having ln(0) as 1/0 makes 0/0=0^0.
Yee says:

February 25, 2013 at 5:54 pm

The Cool Dude,

1.
Index laws can not be abused.

2.
0 can not be a denominator.
The Cool Dude says:

February 26, 2013 at 8:34 pm

Yee,

1. I did not use index laws to say that they were the same thing, I simply made a reference to index laws to show the relation between the two entities. I have already shown innumerable reasons as for why 0^0=0/0 with actual mathematics and real definitions, many of which you have yet to refute, and instead made the mistake of dismissing as irrelevant without any support whatsoever.

2. 0 can be a denominator exactly as much as there can be a root of a negative. They are both non-real answers, the only difference is that one is finite and has an infinitely large plane of relatives, and the other is not finite, with only two other relatives.

You have no mathematical support for 0 not being a denominator.
The only reason anyone has ever claimed division by zero is impossible is summarized by the following:

1*0=0
2*0=0
1*0=2*0
1=2

Reasons the entire idea is a fallacy:
1=2 if and only if 0/0=1 only.
However, if 0/0=1, then
n*0/0=n, and
0/0=n for any n*0=0, which is nearly the exact opposite of only.

The fallacy of association is assuming division by zero is incorrect due to the proof above, which is actually attributed to assuming 0/0=1. In fact, in the situation above, multiplying by zero is as much the problem as dividing by zero.
Yee says:

February 26, 2013 at 11:19 pm

The Cool Dude,

0^0 does not equal 0/0
Bob says:

February 27, 2013 at 2:33 am

Holy cow! 771 Comments!
Anyway, I personally am very convinced that 0^0 equals 1. However, a lot of people that I know, including my Algebra teacher, still says that 0^0 is undefined. Is there any way you can prove that 0^0=1 using elementary algebra (no calculus)?
The Cool Dude says:

February 27, 2013 at 3:34 pm

Yee,
e is defined as
lim (a→∞) of (1+1/a)^a,
And e^x is defined as
lim (a→∞) of (1+x/a)^a

The inverse function, ln(x), is then
lim (a→∞) of (x^(1/a)-1)a
By definition, since plugging the two limits together gets
((x^(1/a)-1)a/a+1)^a
(x^(1/a)-1+1)^a
(x^(1/a))^a
x
Which fulfills the definition of inverse functions.

lim (a→∞) of (x^(1/a)-1)a
Can also be expressed as a limit to zero:
lim (a→0) of (x^a-1)/a

For ln(0), x=0, then
lim (a→0) of (0^a-1)/a=ln(0)
As a is positive:
lim (a→0+) of (0^a-1)/a=(0-1)/a=1/0
As a is negative:
lim (a→0-) of (0^a-1)/a=(1/0-1)/a=(1-0)/(0*a)=1/0
Then ln(0) must be 1/0.

If ln(0)=1/0
e^(1/0)=0
e^(0/0)=0^0

Then with ln(0/0)=ln(0)-ln(0)=0*ln(0)=0/0,
ln(0/0)=0/0
0/0=e^(0/0)
And that 0/0=0^0

Simply saying 0^0≠0/0 over and over does not make it true, and simply ignoring solid math does not make it false.
Noel Evans says:

February 27, 2013 at 5:48 pm

Greetings!

The axioms for the rational number field Q and the real number field R do not allow
zero to have a multiplicative inverse, say 1/0. If it did then 0*1/0 = 1. But other
axioms allow us to prove that 0*x=0 for any x. Hence 0*1/0=0 also. Thus 1=0.
This then makes any x=0 by multiplying both sides of 1=0 by x. Hence all the numbers in the field are equal to zero, so there is only one number in the field. Easy to work with but not very useful.

The usual definition of fractions based on the whole numbers W={0,1,2,3,…} is
F = { a/b | a and b are in W and b is not zero }
Hence writing a/0 for any a in W is not an allowable expression in F. Thus zero for a
denominator is not allowed and so “division by zero” should not even occur. The
fraction symbol “/” is not division.

Division is not an operation given in the field axioms. Division is introduced as multiplication by reciprocals. If neither a nor b is zero then (a/b)*(b/a)=1 so these are reciprocals. Then division can be defined as c/d divided by a/b equals c/d times
b/a where neither a nor b is zero.

Defining exponentiation as x^n = 1*x*x*x*…*x there being n x’s makes x^0=1 for
any x including zero, so 0^0=1 when exponentiation is defined this way.

If we view multiplication as “repeated addition” then multiplication for example 2*x means add x twice. But if we say “add x twice” then we must first “add x once.” But TO WHAT do we add the first x? To get the right answer we must add it to ZERO. That is, 2*x = 0+x+x. Generally n*x=0+x+x+x+…+x there being n x’s.

Similarly if we view exponentiation as “repeated multiplication” then x^2 is “multiplying by x twice.” So we must “multiply by x once” before we can do
it the second time. But WHAT are we multiplying x by the first time? It must be
that we are multiplying ONE by the x the first time. Hence x^2 = 1*x*x. x^1 would
then be multplying 1 by x once. So x^0 would be multiplying 1 by x no times, so we
just have the original 1. And x could be zero, so 0^0=1 using this definition of
exponentiation.

Interesting note: Trying to extend this kind of definition to repeated exponentiation,
for example, x^^3=x^x^x is problematic for several reasons. This “repeated exponentiation” is neither commutative nor associative. Also there is no identity for
the operation in the sense that a^x=x for any x. There is no such a. Since “repeated
exponentiation” is not an operation introduced in the field axioms (as are + and * ) then it would have to be carefully defined and can be, but not in a fashion similar to
the definitions for n*x and x^n above.

Have a blessed day! 🙂
Yee says:

February 27, 2013 at 5:49 pm

Bob,
It’s definition.
It need not prooving.
Yee says:

February 28, 2013 at 6:47 am

The Cool Dude,
0 can not be a denominator.
0^0 does not equal 0/0.
The Cool Dude says:

February 28, 2013 at 4:58 pm

Noel Evans,
You make some good points, but there is one small hole in your logic.

“If we view multiplication as “repeated addition” then multiplication for example 2*x means add x twice. But if we say “add x twice” then we must first “add x once.” But TO WHAT do we add the first x? To get the right answer we must add it to ZERO. That is, 2*x = 0+x+x. Generally n*x=0+x+x+x+…+x there being n x’s.”

This is completely true for all finite numbers. For the question “what is repeated addition added to”, it would in fact be zero, this is due to the fact that zero is the additive identity, and to add something over and over, you must decide which variable, a, solves for a+x=x, where the answer is clearly zero, as zero is the additive identity for all finite numbers, so then, x*0 would be x’s additive identity, because you are not adding anything. Of course, this doesn’t hold for 1/0, as 1/0+5=1/0, but 1/0 is not a finite number, so we will ignore it for now.

“Similarly if we view exponentiation as “repeated multiplication” then x^2 is “multiplying by x twice.” So we must “multiply by x once” before we can do
it the second time. But WHAT are we multiplying x by the first time? It must be
that we are multiplying ONE by the x the first time. Hence x^2 = 1*x*x. x^1 would”

This is of course where the big problem lies. The multiplicative identity would be “a” that solves for a*x=x, which in most conditions is 1, however, just like the additive identity, multiplicative identity has one number that has multiple identities, only this time, the number is real and finite. For x=0, all finite numbers are multiplicative identities, not only one. Being that x^0 would be x’s multiplicative identity, and zero has an infinite number of multiplicative identities, 0^0 has an infinite number of solutions.
Yee says:

February 28, 2013 at 5:58 pm

The Cool Dude,
There is no need to solve such equations.
The Cool Dude says:

March 1, 2013 at 4:34 pm

Yee,
0/0=0^0 for an uncountable collection of different reasons.
Jando says:

March 2, 2013 at 2:09 am

Yee. Admit that you are The Cool Dude. The Cool Dude just be Yee’rself.
Yee says:

March 2, 2013 at 3:04 am

The Cool Dude,
0^0 is different from 0/0.
Noel Evans says:

March 2, 2013 at 1:49 pm

The Cool Dude,

Interesting point about 0 having infinitely many multiplicative identities (0*x=0 for
any x). Perhaps we could call these “local identities for zero.” However, THE
multiplicative identity for a set S is usually an identity simultaneously for ALL the
numbers in S. The axiom would read something like:

There exists an element y in S such that for any x in S, x*y=x. Of course the
usual designation for y is the symbol “1.”

Assuming the system is commutative then we have no need to include y*x=x also. If not commutative then we would need to speak of “right” vs “left” identities.

Also for the 1/0 you mentioned a similar discussion might apply as we could call
the x’s in 1/0 + x = 1/0 local identities for 1/0 where we consider 1/0 as infinity.
I have seen such a system where infinity and minus infinity are bundled together with the reals to have a slightly larger system.

Isn’t it great to be able to discuss such concepts with folks all over the world?

Have a very blessed day! 🙂
The Cool Dude says:

March 2, 2013 at 10:14 pm

Yee,
0^0 is the same as 0/0.
Yee says:

March 3, 2013 at 8:42 am

The Cool Dude,
0^0 is different from 0/0.
0/0 refers to division.
0^0 does not.
The Cool Dude says:

March 3, 2013 at 10:21 pm

Yee,
0^0 is the same as 0/0.
1/i refers to division.
i^3 does not.
Yee says:

March 4, 2013 at 1:18 am

The Cool Dude,
0^0 is different from 0/0.
i^3 does not involve division,
nor does 0^0.
The Cool Dude says:

March 4, 2013 at 5:16 pm

Yee,
0^0 is the same as 0/0.
i^3 doesn’t need division to get the same result,
nor does 0^0.
Bob says:

March 4, 2013 at 10:50 pm

Here is my earlier question rephrased:
How can you prove that 0^0 (zero raised to the zeroth power) equals 1, using only algebra? Furthermore, why isn’t 0^0 undefined? Please don’t use calculus or trigonometry. Thank you.
Yee says:

March 4, 2013 at 11:52 pm

Bob,
0^0 is a matter of definition, which can not be proven.
0^0 is not defined by many mathematicains because continuity is overemphasized.
Yee says:

March 4, 2013 at 11:53 pm

The Cool Dude,
Is 0 the same as 1/(1/0)?
Bob says:

March 5, 2013 at 2:26 am

Oh, and please don’t say something like “it’s definition” or something like that. I need a definition and a proof. Please reply soon. Thanks.
Bob says:

March 5, 2013 at 2:32 am

Again, I agree that, by definition, 0^0=1. I want to prove to my friends and my algebra teacher that 0^0=1. Why isn’t it undefined? Please don’t use calculus.
Don’t think that I am crazy writing math comments at 2:00 am; I live in the Pacific Time zone. I suppose I could say I stay up a little late.
Mr. Mathematician, may I have some help please?
The Cool Dude says:

March 5, 2013 at 6:34 am

Yee,
Absolutely.

Bob,
Using real, imaginary, or complex exponents, if you set 0^0 equal to any number that doesn’t fall into 0^n for any n, you can show that it also equals all other numbers.

If you set 0/0 equal to any number at all, you can show with multiplication and addition that 0/0 is equal to all other numbers.

They are sets, and although, technically contain one, leaving out an infinite number of results is extremely sloppy.
Error: Unable to create directory uploads/2024/11. Is its parent directory writable by the server? The Physicist says:

March 5, 2013 at 10:35 am

@Bob
Absent of context, it is undefined. Often (you can find a few examples higher in this comment thread) it’s more convenient to write a note that says “0^0=1” rather than providing a long explanation, but fundamentally it isn’t a definite value the way “1+1” or “0^5” are definite values.
Yee says:

March 5, 2013 at 10:40 am

Bob,
No,
It is not “by definition, 0^0=1” that you should agree;
it is “0^0=1 is reasonable definition” that you should agree.
It is not a proof that you need;
it is a reason that you need.

0^0=1 is useful, that is the reason.
Yee says:

March 5, 2013 at 6:15 pm

1.
In order to simplify a polynomial a[0]+a[1]*x+…+a[n]*x^n as

n
Σ a[k]*x^k
k=0

x^0 is regarded as 1 for all x.
0^0 must be 1.

2.
Binomial theorem

(x+y)^n=
n
Σ C(n,k)*x^(n-k)*y^k
k=0

If we spread (1-1)^0,
we obtain 0^0=1.
In order to avoid it,
mathematicians set a condition n>0,
that is to cut the head of Pascal’s triangle.
Even so,
if we spread (1+0)^1,
0^0=1 is unavoidable.

3.
In order to satisfy index laws
(0^0)^2=0^(0*2)=0^0
0^0=0^(-0)=1/0^0
0^0=1 is the only solution.

These are examples why 0^0=1 is reasonable definition.
On the contrary, leaving 0^0 undefined is ridiculous,
which is the result of overemphasis of continuity.
Yee says:

March 5, 2013 at 9:01 pm

The Cool Dude,
Is 0 the same as 0^2/0^1?
The Cool Dude says:

March 5, 2013 at 10:25 pm

Yee,
0^0 is 1.
It is just not only 1.

For the same reason, 0^2/0^1 is zero.
It is just not only zero.

Though simplifying 0^0 as either 1 or 0, in either situation, would be technically correct, leaving out an infinite number of alternate solutions is extremely careless, especially when the solution, 0^0, was obtained from a system of equations for which there are extraneous qualifiers.

If, for example, your system was two dimensional, and your equations were:
0^y=5
x=5*x
y=1+xy
Then immediately you would assume the system is impossible, since the first equation makes y=ln(0)^-1, the second equation makes x=0, and the third equation makes x=-ln(0)+1, two results of which do not even exist, and the second two even disagree.

However, if you realize that x=5*x and 1/x=5/x are the exact same system, then you understand that x can also be 1/0. If x=1/0, then the second equation makes y=0.

When plugging the results back in,
0^0=5
1/0=5/0
0=1+0/0
Which, when raising to the power of zero (Or i2π/ln(5)), moving the five to the denominator, and adding one to both sides after putting the first one into the fraction as zero, respectively, ultimately amounts to three statements:
1/0=1/0,
0^0=1
0/0=1
If you assumed 0^0 and/or 0/0 was 1, you would be technically correct, but you would have never received the correct result, and would be left assuming the system was impossible.

Although the subject of the possibility of a system of equations comes into question when one or more variables is not actually real, similar examples can be created where all variables are real, but the solution can still not be gathered by traditional methods. At times, the only suitable replacement for guessing is moving through values of nonexistent numbers.

Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

1,177 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

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